I imagine this is already found but I cannot find the proof.
The formula also works for $f_1=\sqrt{2}^{1/x}$ to get many multiples of $\pi$
Can you also prove that when $f_1=\sqrt{\frac{2}{\pi}}$ then $f_n$ tends to $\sqrt{n}-.5$ ?
Example for $f_1=2$
Example for $f_1=\sqrt{\frac{2}{\pi}}$
On
The question is about a slight tweak of the Fibonacci recursion.
Let the sequence $\,f_n\,$ be defined by the recursion
$$ f_n = 1/f_{n-1} + f_{n-2}. \tag1 $$
Suppose the initial values of the sequence are
$$ f_0 = x/y, \quad f_1 = y. \tag2 $$
The next few sequence terms are
$$ f_2 = \frac{1+x}y, f_3 = \frac{y(2+x)}{1+x}, f_4 = \frac{(1+x)(3+x)}{y(2+x)}, f_5 = \frac{y(2+x)(4+x)}{(1+x)(3+x)}. \tag3 $$
The numerators and denominators can be written in terms of the Pochhammer symbol or rising factorial
$$ x^{(n)} := x(x+1)(x+2)\cdots(x+n-1) \tag4 $$
by the formulas
$$ f_{2n} = \frac2y\frac{((x+1)/2)^{(n)}}{((x+2)/2)^{(n-1)}}, \qquad f_{2n+1} = y\frac{((x+2)/2)^{(n)}}{((x+1)/2)^{(n)}}. \tag5 $$
Notice the simple result that
$$ \frac{f_{n+2}}{f_n} = \frac{n+1+x}{n+x}. \tag6 $$
Equation $(5)$ implies that
$$ \frac{f_{2n+1}}{f_{2n}} = \frac{4y^2}{x^2(x+2n)} \left(\frac{(x/2)^{(n+1)}}{((x+1)/2)^{(n)}}\right)^2. \tag7 $$
In the case that $x=0$ this simplifies to
$$ \frac{f_{2n+1}}{f_{2n}} = \frac{y^2n!^4 2^{4n}}{(2n)(2n)!^2}. \tag8 $$
Use Stirling's formula for $n!$ to show $\lim_{n\to\infty} \frac{f_{2n+1}}{f_{2n}} =y^2\frac{\pi}2 $ which is $2\pi$ if $y=2.$
Now use equation $(6)$ to get $\lim_{n\to\infty} \frac{f_{n+2}}{f_n} = 1$ and thus, $\lim_{n\to\infty} \frac{f_{2n}}{f_{2n-1}} =(2\pi)^{-1}.$
This answer the question
Can you prove that $\frac{f_n}{f_{n-1}}$ converges to $2\pi$ and $\frac{1}{2\pi}$ if $f_n=\frac{1}{f_{n-1}}+f_{n-2}$ where $f_0=0$ and $f_1=2$?
Now one answer to the other question
Can you also prove that when $f_1=\sqrt{\frac{2}{\pi}}$ then $f_n$ tends to $\sqrt{n}-.5$ ?
is to use the method of undetermined coefficients. The two limits are equal if and only if $y = f_1 = \sqrt{\frac{2}{\pi}}$. In that case, use the Ansatz
$$ {f_n}^2 = n + a_0 + a_1n^{-1} + a_2n^{-2} + \dots \tag9 $$
which must agree with equation $(6)$ using $x=0,$ which condition
$$ \left(\frac{f_{n+2}}{f_n}\right)^2 = \frac{(n+1)^2}{n^2} = 1 + 2n^{-1} + n^{-2} \tag{10} $$
uniquely determines the coefficients in equation $(9)$ to get
$$ {f_n}^2 = n - \frac12 + \frac18 n^{-1} + \frac1{16}n^{-2} - \frac{5}{128}n^{-3} - \frac{23}{256}n^{-4} + \dots. \tag{11} $$
Take the square root to get
$$ f_n = \sqrt{n}\left(1 - \frac14 n^{-1} + \frac1{32}n^{-2} +\frac5{128}n^{-3} -\frac{21}{2048}n^{-4} + \dots \right). \tag{12} $$
Thus,
$$ {f_n}^2 \approx n - \frac12, \qquad f_n \approx \sqrt{n} - \frac14 \frac1{\sqrt{n}}. \tag{13} $$
Leave $f_1$ arbitrary; $f_{n}f_{n-1}=f_{n-1}f_{n-2}+1$ implies $f_{n}f_{n+1}=n$ (by induction). Let $$ f_n=\frac{\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(\frac n2\right)}a_n; $$ then $a_{n}a_{n+1}=2$, thus $a_{2n+1}=a_1$ and $a_{2n+2}=2/a_1$ for all $n$. Now the known limit $$ \lim_{x\to+\infty}\frac{\Gamma(x+a)}{x^a\ \Gamma(x)}=1\qquad(a\in\mathbb{R}) $$ easily gives $\displaystyle\color{blue}{\lim_{n\to\infty}\frac{f_{2n+1}}{\sqrt n}}=a_1\color{blue}{=f_1\sqrt\pi}$ and $\displaystyle\color{blue}{\lim_{n\to\infty}\frac{f_{2n}}{\sqrt n}}=\frac2{a_1}\color{blue}{=\frac2{f_1\sqrt\pi}}$.
(An elementary alternative here is to use the Wallis product.)
For $f_1=\sqrt{2/\pi}$ we get $a_n=\sqrt2$ identically, and the asymptotics of $$ f_n=\frac{n}{\sqrt{2\pi}}\mathrm{B}\left(\frac{n+1}2,\frac12\right)=n\sqrt\frac2\pi\int_0^1\frac{x^n\,dx}{\sqrt{1-x^2}}=n\sqrt\frac2\pi\int_0^\infty\frac{e^{-nt}\,dt}{\sqrt{e^{2t}-1}} $$ can be obtained using Watson's lemma. The (correct) result is $$ f_n\asymp n^{1/2}\color{red}{-\frac14 n^{-1/2}}\color{gray}{+\frac1{32}n^{-3/2}+\frac5{128}n^{-5/2}+\dots} $$